Mathematical Methods for Physics and Engineering - Matematica.NET

PDES: SEPARATION OF VARIABLES AND OTHER METHODSsolution is immediate:ū(x, s) =A exp(√ ) ( √ )s sκ x + B exp −κ x ,where the constants A and B may depend on s.We require u(x, t) → 0asx →∞and so we must also have ū(∞,s) = 0; consequentlywe require that A = 0. The value of B is determined by the need for u(0,t)=u 0 and hencethat∫ ∞ū(0,s)= u 0 exp(−st) dt = u 00s .We thus conclude that the appropriate expression for the Laplace transform of u(x, t) isū(x, s) = u ( √ )0 ss exp −κ x . (21.72)To obtain u(x, t) from this result requires the inversion of this transform – a task that isgenerally difficult and requires a contour integration. This is discussed in chapter 24, butfor completeness we note that the solution is( )] xu(x, t) =u 0[1 − erf √ ,4κtwhere erf(x) is the error function discussed in the Appendix. (The more complete sets ofmathematical tables list this inverse Laplace transform.)In the present problem, however, an alternative method is available. Let w(t) betheamount of salt that has diffused into the tube in time t; thenand its transform is given by¯w(s) ===w(t) =∫ ∞0∫ ∞0∫ ∞0∫ ∞0dt exp(−st)dx∫ ∞0ū(x, s) dx.u(x, t) dx,∫ ∞0u(x, t) dxu(x, t)exp(−st) dtSubstituting for ū(x, s) from (21.72) into the last integral and integrating, we obtain¯w(s) =u 0 κ 1/2 s −3/2 .This expression is much simpler to invert, and referring to the table of standard Laplacetransforms (table 13.1) we findw(t) =2(κ/π) 1/2 u 0 t 1/2 ,which is thus the required expression for the amount of diffused salt at time t. ◭The above example shows that in some circumstances the use of a Laplacetransformation can greatly simplify the solution of a PDE. However, it will havebeen observed that (as with ODEs) the easy elimination of some derivatives isusually paid for by the introduction of a difficult inverse transformation. Thisproblem, although still present, is less severe for Fourier transformations.748